A particle of mass 1kg is subjected to a force which depends on the...

Given here: \overset{\to}{F} = - k\left(x\overset{^}{i} + y\overset{^}{j}\right)kgm{s}^{- 2} and $m = 1kg$

In x-direction, ${F}_{x} = - x = m{a}_{x}$

So, acceleration, ${a}_{x} = \dfrac{{d}^{2}x}{d{t}^{2}} = - x$

Now, for particle executing SHM, displacement along x-direction is

$⇒ x = {A}_{x}\sin \left(\omegat + {ϕ}_{x}\right)$, here, angular frequency, $\omega = 1rad{s}^{- 1}$ and velocity, ${v}_{x} = {A}_{x}\omega cos\left(\omegat + {ϕ}_{x}\right)$

Given at $t = 0, x = \dfrac{1}{\sqrt{2}}m$ and ${v}_{x} = - \sqrt{2}m{s}^{- 1}$

So, putting the values, we get

$\dfrac{1}{\sqrt{2}} = {A}_{x}sin{ϕ}_{x}$ and $- \sqrt{2} = {A}_{x}cos{ϕ}_{x}$

From above two equations,

$⇒ tan{ϕ}_{x} = - \dfrac{1}{2} . . . \left(1\right)$

And ${A}_{x} = \sqrt{\dfrac{5}{2}}m . . . \left(2\right)$

Similarly, along y-direction,

${F}_{y} = - y = m{a}_{y}$,

$⇒ {a}_{y} = \dfrac{{d}^{2}y}{d{t}^{2}} = - y$

So, displacement, $y = {A}_{y}\sin \left(\omegat + {ϕ}_{y}\right)$ and velocity ${v}_{y} = {A}_{y}\omega cos\left(\omegat + {ϕ}_{y}\right)$

Given at $t = 0, y = \sqrt{2}m$ and ${v}_{y} = \sqrt{2}m{s}^{- 1}$

So, putting the values, we get

$\sqrt{2} = {A}_{y}sin{ϕ}_{y}$ and $\sqrt{2} = {A}_{y}cos{ϕ}_{y}$

From above two relations, we have

$⇒ {ϕ}_{y} = \dfrac{\pi}{4} . . . \left(3\right)$ and ${A}_{y} = 2m . . . \left(4\right)$

Now, the value of \left(x{v}_{y} - y{v}_{x}\right) = \sqrt{\dfrac{5}{2}}\sin \left(\omegat + {ϕ}_{x}\right) \times 2\cos \left(\omegat + {ϕ}_{y}\right) - 2\sin \left(\omegat + {ϕ}_{y}\right) \times \sqrt{\dfrac{5}{2}}\cos \left(\omegat + {ϕ}_{x}\right)$</p><p>$= \sqrt{\dfrac{5}{2}} \times 2\left(\sin \left(\omegat + {ϕ}_{x}\right)\cos \left(\omegat + {ϕ}_{y}\right) - \sin \left(\omegat + {ϕ}_{y}\right) \times \cos \left(\omegat + {ϕ}_{x}\right)\right)

$= \sqrt{10}\sin \left({ϕ}_{x} - {ϕ}_{y}\right)$

$= \sqrt{10}\left(sin{ϕ}_{x}cos{ϕ}_{y} - cos{ϕ}_{x}sin{ϕ}_{y}\right)$

$= \sqrt{10}\left(\dfrac{1}{\sqrt{5}} \times \dfrac{1}{\sqrt{2}} - \left(- \dfrac{2}{\sqrt{5}}\right) \times \dfrac{1}{\sqrt{2}}\right)$

= 3